# Deriving time dilation from line element

I am having trouble deriving the time dilation. I am using $$(-, +, +, +)$$ sign convention.

For Minkowski metric, the line element is equal to: $$ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2$$

For a motionless ($$dx = dy = dz = 0$$) distant observer the line element is equal to: $$ds^2 = -c^2dt^2$$

And for a moving object the line element is equal to: $$ds^2 = -c^2d\tau^2 + dx^2 + dy^2 + dz^2$$ Where $$\tau$$ is its proper time.

Now, as far as I understand, I need to set these two line elements equal and solve for $$\frac{dt}{d\tau}$$: $$-c^2dt^2 = -c^2d\tau^2 + dx^2 + dy^2 + dz^2$$ $$\left(\frac{dt}{d\tau}\right)^2 = 1 - \frac{\left(\frac{dx}{d\tau}\right)^2 + \left(\frac{dy}{d\tau}\right)^2 + \left(\frac{dz}{d\tau}\right)^2}{c^2}$$ $$\frac{dt}{d\tau} = \sqrt{1 - \frac{v^2}{c^2}}$$

This does not make sense to me, since the time measured by the motionless stationary observer $$t$$ should be greater than the proper time $$\tau$$. So $$\frac{dt}{d\tau} > 1$$. This is not true for $$\frac{dt}{d\tau} = \sqrt{1 - \frac{v^2}{c^2}}$$. I already saw $$\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$ somewhere, but I'm not sure how to arrive at it.

For Schwarzschild's metric, I am solving this for $$\frac{dt}{d\tau}$$: $$-c^2 dt^2 = -\left(1 - \frac{r_s}{r}\right)c^2 d\tau^2 + \left(1 - \frac{r_2}{r}\right)^{-1} dr^2 + r^{2} \left(d\theta^2 + \sin^2 \theta \, d\phi^2\right)$$

Doing this, I arrive at: $$\left(\frac{dt}{d\tau}\right)^2 = \left(1 - \frac{r_s}{r}\right) - \left(1 - \frac{r_2}{r}\right)^{-1} \frac{1}{c^2} \left(\frac{dr}{d\tau}\right)^2 - \left(\frac{r}{c}\right)^{2} \left(\left(\frac{d\theta}{d\tau}\right)^2 + \sin^2 \theta \, \left(\frac{d\phi}{d\tau}\right)^2\right)$$ This doesn't seem even close to being correct. I already saw other posts and the equation is very different. I want to have it universal, so I can plug in any values. Where am I making a mistake? Maybe the change in position over a change in time (e.g. $$\frac{dx}{d\tau}$$) should be with respect to $$t$$ instead ($$\frac{dx}{dt})$$?

• Your initial line element is incorrect, see physics.stackexchange.com/questions/54569/… Commented Nov 3, 2023 at 8:38
• @Eletie I'm sorry, could you point out the mistake to me? I got it from en.m.wikipedia.org/wiki/Line_element#Minkowskian_spacetime and en.m.wikipedia.org/wiki/… Both say $ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2$ Commented Nov 3, 2023 at 9:06
• check your second and third equations... they might not be adequately expressed. Commented Nov 3, 2023 at 12:54
• I can't make sense of the third equation or your equating of the ${\rm d}s^2$. I would recommend taking a step back and thinking about what $t$, $s$, and $\tau$ are measuring. Commented Nov 3, 2023 at 15:40

For Minkowski metric, the line element is equal to $$ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2. \tag{1}$$ In a rest coordinate system ($$dx = dy = dz = 0$$) time $$t$$ is equal to proper time $$\tau$$ $$ds^2 = -c^2d\tau^2. \tag{2}$$ Equating equations $$(1)$$ and $$(2)$$ results in $$-c^2 d\tau^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2, \tag{2}$$ $$\left(\frac{d\tau}{dt}\right)^2 = 1 - \frac{1}{c^2}\left(\Big(\frac{dx}{dt}\Big)^2 + \Big(\frac{dy}{dt}\Big)^2 + \Big(\frac{dz}{dt}\Big)^2\right), \tag{3}$$ and finally in correct equation for the dilatation time $$\frac{d\tau}{dt} = \sqrt{1 - \frac{v^2}{c^2}}. \tag{4}$$

• I think I'm missing something. Is $t$ the time of the motionless clock and $\tau$ the proper time of the moving clock? Or is it the opposite? What are the general steps for any metric? Commented Nov 3, 2023 at 17:13
• @Yachim A clock in a rest frame measures proper time. $t$ is the so-called coordinate time from the right side of eq. (1). Moving is a relative notion. My clock rests other clock moves in reference to me. From the viewpoint of other clock you can say the same. About general steps for any metric I cannot answer you yet. I have to think about first.
– JanG
Commented Nov 3, 2023 at 19:22
• The sign of $1/c^2$ in $(3)$ should be positive (or $t$ and $\tau$ should swap places). Commented Nov 4, 2023 at 1:14
• @KrisWalker You are right!
– JanG
Commented Nov 4, 2023 at 7:37
• @Yachim I have had to correct my equation (3), sorry! Now, you have three correct answers.
– JanG
Commented Nov 4, 2023 at 8:03

The problem appears to be in your initial setup.

For Minkowski metric, the line element is equal to: $$ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2$$

That is correct.

For a motionless ($$dx = dy = dz = 0$$) distant observer the line element is equal to: $$ds^2 = -c^2dt^2$$

That is correct but not relevant. The time dilation formula is not restricted to distant observers. Time dilation is the ratio between coordinate time and proper time for a given clock.

And for a moving object the line element is equal to: $$ds^2 = -c^2d\tau^2 + dx^2 + dy^2 + dz^2$$ Where $$\tau$$ is its proper time.

This one is incorrect. The correct expression is $$ds^2 = -c^2d\tau^2$$

Deriving time dilation is straightforward. You simply start with the line element, substitute this corrected expression, divide both sides by $$-c^2 dt^2$$ and simplify. The result is $$1/\gamma^2$$, where $$\gamma$$ is the time dilation factor.

$$ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2$$$$-c^2d\tau^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2$$$$\frac{d\tau^2 }{dt^2} = 1 -\frac{ dx^2 }{c^2 dt^2} -\frac{ dy^2 }{c^2 dt^2}-\frac{ dz^2 }{c^2 dt^2}$$$$\frac{1}{\gamma^2} =1-\frac{v^2}{c^2}$$

Consider a frame in which the coordinates are $$(t, x, y, z)$$. An object moving in this frame will have some trajectory with line element $${\rm d}s^2=-c^2{\rm d}t^2+{\rm d}x^2+{\rm d}y^2+{\rm d}z^2.\tag*{(1)}$$ Now boost to the frame of the object, in which the coordinates are $$(t^\prime,x^\prime,y^\prime,z^\prime)$$. Since the object is at rest in this frame, we have that $${\rm d}x^\prime={\rm d}y^\prime={\rm d}z^\prime=0$$, making the line element $${\rm d}s^2=-c^2{\rm d}{t^\prime}^2.\tag*{(2)}$$ This is an invariant, so we recognize $$t^\prime$$ as an invariant "proper time" of the object and typically denote it as $$\tau$$, so that $${\rm d}s^2=-c^2{\rm d}\tau^2$$. And since it is an invariant, $$(1)$$ and $$(2)$$ are equal, so we have that $$-c^2{\rm d}\tau^2=-c^2{\rm d}t^2+{\rm d}x^2+{\rm d}y^2+{\rm d}z^2,\tag*{(3)}$$ which is easily rearranged to obtain $${\rm d}t/{\rm d}\tau=1/\sqrt{1-\boldsymbol{v}\cdot\boldsymbol{v}/c^2}$$, where $$\boldsymbol{v}={\rm d}\boldsymbol{x}/{\rm d}t$$.